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Help with some Lebesgue measure problems


john33

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Hi everyone, I'm trying to solve some Lebesgue problems from my exercise book and I got stuck in some of them:

  • Prove that if a set A has zero measure, then its interior is empty. I've thinking on suppose the contrary and find an open subset of A with positive measure, but I'm not really sure if it's the right way.
  • True or false: f is integrable if and only if |f| is integrable over Rn. I know that if f is measurable, then it's correct, but here there is no previous condition so I don't know if the statement is true or false.

Could you give me some tips to solve it? Thanks in advance.

Edited by john33
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55 minutes ago, john33 said:

Hi everyone, I'm trying to solve some Lebesgue problems from my exercise book and I got stuck in some of them:

  • Prove that if a set A has zero measure, then its interior is empty. I've thinking on suppose the contrary and find an open subset of A with positive measure, but I'm not really sure if it's the right way.
  • True or false: f is integrable if and only if |f| is integrable over Rn. I know that if f is measurable, then it's correct, but here there is no previous condition so I don't know if the statement is true or false.

Could you give me some tips to solve it? Thanks in advance.

For the first one I would try the inverse direction; something like (as @mathematicsuggests),

 

intCØμ(C)0

Think open balls.

I see you already have good help, so I'll leave it at that.

Cheers.

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  • 3 months later...
On 4/8/2021 at 3:33 PM, mathematic said:

Non-empty open sts must have positive measure, so your approach to the first question is correct.  The second question is almost from the definition.

The first statement is  not true.  For example, a singleton set has 0 measure.  What is true that any set containing an interval has non-zero measure.

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53 minutes ago, Country Boy said:

The first statement is  not true.  For example, a singleton set has 0 measure.  What is true that any set containing an interval has non-zero measure.

Usual definition of open - singleton is NOT open.  Every neighborhood contains points outside the set.

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